ANSWER
![\begin{equation*} 468.75\text{ foot-pound} \end{equation*}](https://img.qammunity.org/2023/formulas/physics/college/tgb3zfdsuythz9s05ztq63nr18qgsqpldc.png)
Step-by-step explanation
Let the bucket be lifted x feet.
The weight of the bucket at x feet is given by:
![20-0.1x](https://img.qammunity.org/2023/formulas/physics/college/eemhefikp4qh9x7slwvlxw99koy2fu2w44.png)
The work done in lifting the bucket by dx feet is:
![dW=(20-0.1x)dx](https://img.qammunity.org/2023/formulas/physics/college/6bqg10q9z1g0hqe8tna3vlvtvw8ybmpa0f.png)
The total work done is the integral of the work done in lifting the bucket x feet, that is:
![W=\int(20-0.1x)dx](https://img.qammunity.org/2023/formulas/physics/college/np9zi5h751s5dqce7wzssqq3o9s1qkf295.png)
Hence, the work done in lifting the bucket 25feet is:
![\begin{gathered} W=\int_0^(25)(20-0.1x)dx \\ \\ W=(20x-(0.1x^2)/(2))_0^(25) \\ \\ W=(20*25-(0.1*25^2)/(2))-(20*0-(0.1*0^2)/(2)) \\ \\ W=500-31.25 \\ \\ W=468.75\text{ foot-pound} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/s66sjl9l37pzpov8aqy8aipnblmt97hs3s.png)
That is the answer.